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Paul Erdős gave a simple argument to show that $\pi(x) \ge \dfrac{1}{2}\log_2 x$.

Is it possible to tweak the argument and get a better estimate? I am wondering how good an estimate for $\pi(x)$ can be achieved using a variation of his reasoning.

I explored the possibility of changing $m^2$ to $m^3$ so that we have for any $y \le x$, we have:

$y = p_1^{e_1}p_2^{e_2}\ldots p_n^{e_n}m^3$ and $e_i\in\left\{0,1,2\right\}$ and $m \in \mathbb{Z}$

But this gets us to: $m \le \sqrt[3]{x}$ so that $3^n\times \sqrt[3]{x} \ge x$ and $\pi(x) \ge \dfrac{2}{3}\log_3 x$ which is weaker than $\pi(x) \ge \dfrac{1}{2}\log_2 x$

Has anyone thought of other creative tweaks that can improve the result?

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Larry Freeman
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    Considering the research that has gone into approximations of $\pi(x)$ it is no surprise that an approximation $\hat{\pi}(x)$ and a factor $(1-\epsilon)$ give a lower bound $\hat{\pi}(1-\epsilon)$ that is better than something derived for another purpose. Even Dusart's weaker bounds (Wiki) are much better than the one here. Nice idea though. – daniel Jan 24 '15 at 10:00

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The same idea also gives $p_n\leq 4^n$ and a rather trivial improvement of that is:

$n\in\Bbb Z_+,\;1\leq x\leq p_n\implies x=p_1^{e_1}\cdots p_n^{e_n}\cdot m^2,\;e_i=0,1\wedge\;m^2<p_n$.
If $e_n=1$ then $0=e_1=\cdots=e_{n-1}$ and if $e_n=0$ there is at most $2^{n-1}$ ways to chose $x$. Similar to Erdős proof $p_n\leq(2^{n-1}+1)\sqrt p_n$ which gives $p_n\leq (2^{n-1}+1)^2$.

A challenge would be to prove $p_n\leq 2^{n-1}+1$ without using the prime number theorem.

Lehs
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  • Isn't it very easy to prove $p_n < 2^{n-2}$ using Bertrand's Postulate? $p_6 = 13 < 2^{6-2}$ Assume up to $n$ $p_n < 2^{n-2}$ Then, $p_{n+1} < 2\cdot p_n < 2^{(n+1)-2}$. Am I missing something? – Larry Freeman Feb 02 '15 at 06:06
  • @LarryFreeman: I assumed that Bertrand's postulate was proved with PNT. The idea, I thought, was to use only very elementary tricks, like Paul Erdős did. – Lehs Feb 02 '15 at 09:35
  • Thanks for explaining. Bertrand's Postulate does not require PNT. A recent proof removes Stirling's Approxmation. I agree that it would be very nice to have an even simpler proof establishing $p_n \le 2^{n-1}+1$ – Larry Freeman Feb 02 '15 at 09:39